F.4 Amaths Question

F.4 Amaths Question
(a) If sin(θ+φ)=ksin(θ-φ), where cos θ and cos φ ≠0, show that
(k-1)tan θ = (k+1)tan φ.

For sin(θ+φ)=ksin(θ-φ), where cos θ and cos φ ≠0
Dividing LHS by RHS, we get:
sin(θ+φ)/ sin(θ-φ) = k
[sinθcosφ + cosθsinφ] / [sinθcosφ - cosθsinφ] = k
[tanθ/tanφ + 1 ] / [tanθ/tanφ – 1] = k
[tanθ+ tanφ] / [tanθ–tanφ] = k
[tanθ+ tanφ] = k[tanθ–tanφ]
[tanθ+ tanφ] = ktanθ–ktanφ
Ktanθ - tanθ = tanφ + ktanφ
(k-1)tan θ = (k+1)tan φ.


(b) Given that 2θ+φ=90°. Using the result of (a), solve the equation
sin(θ+φ)=2sin(θ-φ) for θ and φ where 0°≦θ, φ≦180°.

Using the above result for sin(θ+φ)=2sin(θ-φ), we have:
(2-1)tanθ = (2+1)tanφ
tanθ = 3tanφ
For 2θ+φ=90°, we get
θ= 45° - 0.5φ
tan(45° - 0.5φ) = 3tanφ
Since tan(A-B)=(tanA-tanB)/(1+tanAtanB), we have
(tan45° - tan0.5φ) / ( 1+ tan45°tan0.5φ) = 3tanφ
(1 - tan0.5φ) / 1 + tan0.5φ) = 3tanφ
Since tan2A=2tanA/(1-tan^2A)
(1 - tan0.5φ) / (1 + tan0.5φ) = 3*2tan0.5φ / (1 – tan^20.5φ)
(1 - tan0.5φ) / (1 + tan0.5φ) = 6tan0.5φ / (1 - tan0.5φ) * (1 + tan0.5φ)
1 - 2 tan0.5φ + tan^20.5φ = 6tan0.5φ
1 - 8 tan0.5φ + tan^20.5φ = 0
Solving the quadratic equation

tan0.5φ = {-(-5 ) + sqrt[ (-5)^2 – 4*1*1]} / 2
tan0.5φ = 7.87

0.5φ = arctan7.87
0.5φ = 82.76°
φ = 165.5°
and θ = 45° - 165.5°
θ = 45° - 165.5°
= -120.5°
= 360 -120.5°
= 239.5°

Since 0°≦θ and φ≦180°,
this set of data is rejected.

the another root of the quadratic equation is
tan0.5φ = {-(-5 ) - sqrt[ (-5)^2 – 4*1*1]} / 2
tan0.5φ = 0.127

0.5φ = 7.24°
φ = 14.5°
and θ = 45° - 7.24°
θ = 45° - 7.24°
= 37.8°

This set of data satisfies the condition that 0°≦θ and φ≦180°

Ans: φ = 14.5° and θ = 37.8°

(b)

sin(θ+φ)=2sin(θ-φ)
tan θ=3 tan φ
tan θ=3 tan(90°- 2θ)
tan θ=3 / tan(2θ)
tan θ=3 (1- tan^2 θ) / 2tanθ
2 tan^2 θ = 3 - 3 tan^2 θ
tan^2 θ = 3/5
....

2007-06-21 20:16:48 補充：
While it is correct, it overcomplicates. Isn't my answer more simple and straightforward?